What are the solutions to the equation z^2+4z-45=0 ?

The solutions to the equation z^2+4z-45=0 are z=5,z=-9

Find the zeros of z^2+4z-45=0

The zeros of z^2+4z-45=0 are z=5,z=-9

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z^{2} + 4 z - 45 = 0

z = 5, z = - 9

Solution steps

z^{2} + 4 z - 45 = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 45 )}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot (- 45) = 14

z_{1, 2} = \frac{- 4 \pm 14}{2 \cdot 1}

Separate the solutions

z_{1} = \frac{- 4 + 14}{2 \cdot 1}, z_{2} = \frac{- 4 - 14}{2 \cdot 1}

z = \frac{- 4 + 14}{2 \cdot 1} : 5

z = \frac{- 4 - 14}{2 \cdot 1} : - 9

The solutions to the quadratic equation are:

z = 5, z = - 9

\frac{3 h}{4} = 9

\frac{28}{7}

23 + 3

(- 3 x y^{2})^{4}

∣ - 6 ∣ + 9

What are the solutions to the equation z^2+4z-45=0 ?
The solutions to the equation z^2+4z-45=0 are z=5,z=-9
Find the zeros of z^2+4z-45=0
The zeros of z^2+4z-45=0 are z=5,z=-9